If 1900 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
so i tried doing this
v=x^2y
x^2+4xy=1900
y=(1900-x^2)/4x
=475x-x/4

V=x^2(475x-x/4)
taking the derivative and setting it to 0 i get
12.58 but thats not right
can anyone see what i did wrong

August 9th 2010, 04:36 PM

eumyang

Quote:

If 1900 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
so i tried doing this
v=x^2y
x^2+4xy=1900
y=(1900-x^2)/4x
=475x-x/4

The last step above is wrong.

So

August 9th 2010, 04:46 PM

TKHunny

Unfortunately, you seem to have confused yourself. I will give you a great gift, My Friend. It is the gift of formality. You have gravely underestimated the value of being careful, of being explicit, and of being complete.

1) You switched from 'v' to "V" in the middle of your demonstration. This may be only a typo, but it is confusing.
2) You defined neither 'x' not 'y'. You should have done this carefully at the beginning. It takes very little time, but it will save you often.

x = length of each side of the square base.
y = height of the box.

3) You did not explain your steps. Your second was very confusing. I first thought you were trying to find a derivative of v(x,y). It took a while to see what you were doing. WRITE IT DOWN and no one will have to be confused.

In this case, however, you are not done. Are you SURE it's a maximum? REALLY SURE? Derivatives don't know the difference between minima and maxima. Derivatives don't work at endpoints, either. There is SO MUCH to think about. There just isn't time to be lazy or sloppy.